Factorize the given equation : $8 a^{3} + 27 b^{3} + 64 c^{3} - 72 a b c$ . And, also find what factors does this equation have?

(2 a + 3 b + 4 c) (4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b - 12 b c - 8 a c)

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(a + 3 b + 4) (4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b - 12 c - 8 a)

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(3 b + 4 c) (4 a^{2} + 9 b^{2} + c^{2} - 6 a b - 12 c - 8 a c)

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(a + 3 b + c) (a^{2} + b^{2} + 16 c^{2} - a b - 2 b c - c)

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Solution

The correct option is A $(2 a + 3 b + 4 c) (4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b - 12 b c - 8 a c)$
$8 a^{3} + 27 b^{3} + 64 c^{3} - 72 a b c$
$= 2^{3} a^{3} + 3^{3} b^{3} + 4^{3} c^{3} - 3 (2 a) (3 b) (4 c)$
$= (2 a)^{3} + (3 b)^{3} + (4 c)^{3} - 3 (2 a) (3 b) (4 c)$
Using, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - a c - b c)$ , we get
$(2 a)^{3} + (3 b)^{3} + (4 c)^{3} - 3 (2 a) (3 b) (4 c)$
$= (2 a + 3 b + 4 c) ((2 a)^{2} + (3 b)^{2} + (4 c)^{2} - (2 a) (3 b) - (3 b) (4 c) - (2 a) (4 c))$
$= (2 a + 3 b + 4 c) (4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b - 12 b c - 8 a c)$
$∴ 8 a^{3} + 27 b^{3} + 64 c^{3} - 72 a b c = (2 a + 3 b + 4 c) (4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b - 12 b c - 8 a c)$
The factors of the given equation are $(2 a + 3 b + 4 c)$ and $(4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b - 12 b c - 8 a c)$

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